Mathematics, 10.03.2020 07:57 hotrdks946

HELP PLEASE

problems 11 through 17

HELP PLEASE problems 11 through 17

Answers: 2

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Answers

Answer from: genyjoannerubiera

Part 11) The vertex is the point (4,32)

Part 12) The vertex is the point (1,-5)

Part 13) The vertex is the point (-2,5)

Part 14) The vertex is the point (-1,-1)

Part 15) The vertex is the point (1,8)

Part 16) The vertex is the point (3,-26)

Part 17) The vertex is the point (-5,-32)

Step-by-step explanation:

Part 11) we have

y=-x^2+8x+16

we know that

The quadratic equation written in vertex form is equal to

y=a(x-h)^2+k

where

a is the leading coefficient

(h,k) is the vertex

Convert to vertex form

y=-x^2+8x+16

Factor -1

y=-(x^2-8x)+16

Complete the square

y=-(x^2-8x+16)+16+16

y=-(x^2-8x+16)+32

Rewrite as perfect squares

y=-(x-4)^2+32

The vertex is the point (4,32)

Part 12) we have

y=3x^2-6x-2

we know that

The quadratic equation written in vertex form is equal to

y=a(x-h)^2+k

where

a is the leading coefficient

(h,k) is the vertex

Convert to vertex form

y=3x^2-6x-2

Factor 3

y=3(x^2-2x)-2

Complete the square

y=3(x^2-2x+1)-2-3

y=3(x^2-2x+1)-5

Rewrite as perfect squares

y=3(x-1)^2-5

The vertex is the point (1,-5)

Part 13) we have

y=-2x^2-8x-3

we know that

The quadratic equation written in vertex form is equal to

y=a(x-h)^2+k

where

a is the leading coefficient

(h,k) is the vertex

Convert to vertex form

y=-2x^2-8x-3

Factor -2

y=-2(x^2+4x)-3

Complete the square

y=-2(x^2+4x+4)-3+8

y=-2(x^2+4x+4)+5

Rewrite as perfect squares

y=-2(x+2)^2+5

The vertex is the point (-2,5)

Part 14) we have

y=2x^2+4x+1

we know that

The quadratic equation written in vertex form is equal to

y=a(x-h)^2+k

where

a is the leading coefficient

(h,k) is the vertex

Convert to vertex form

y=2x^2+4x+1

Factor 2

y=2(x^2+2x)+1

Complete the square

y=2(x^2+2x+1)+1-2

y=2(x^2+2x+1)-1

Rewrite as perfect squares

y=2(x+1)^2-1

The vertex is the point (-1,-1)

Part 15) we have

y=-5x^2+10x+3

we know that

The quadratic equation written in vertex form is equal to

y=a(x-h)^2+k

where

a is the leading coefficient

(h,k) is the vertex

Convert to vertex form

y=-5x^2+10x+3

Factor -5

y=-5(x^2-2x)+3

Complete the square

y=-5(x^2-2x+1)+3+5

y=-5(x^2-2x+1)+8

Rewrite as perfect squares

y=-5(x-1)^2+8

The vertex is the point (1,8)

Part 16) we have

y=3x^2-18x+1

we know that

The quadratic equation written in vertex form is equal to

y=a(x-h)^2+k

where

a is the leading coefficient

(h,k) is the vertex

Convert to vertex form

y=3x^2-18x+1

Factor 3

y=3(x^2-6x)+1

Complete the square

y=3(x^2-6x+9)+1-27

y=3(x^2-6x+9)-26

Rewrite as perfect squares

y=3(x-3)^2-26

The vertex is the point (3,-26)

Part 17) we have

y=x^2+10x-7

we know that

The quadratic equation written in vertex form is equal to

y=a(x-h)^2+k

where

a is the leading coefficient

(h,k) is the vertex

Convert to vertex form

y=x^2+10x-7

Complete the square

y=(x^2+10x+25)-7-25

y=(x^2+10x+25)-32

Rewrite as perfect squares

y=(x+5)^2-32

The vertex is the point (-5,-32)

Answer from: Quest

b. 6

step-by-step explanation:

we are given the table representing an exponential function.

we can see from the table that, as x is increasing, the value of y is increasing in the power of 6.

i.e. the following relation is obtained,

x = 1, y = 6

x = 2, y = 36 = $6^{2}$

x = 3, y = 216 = $6^{3}$

x = 4, y = 1296 = $6^{4}$

x = 5, y = 7776 = $6^{5}$

so, the geometric series obtained is $6, 6^{2}, 6^{3}, 6^{4}, 6^{5}$ .

thus, the common ratio is given by $\frac{6^{2}}{6}$ = $\frac{36}{6}$ = 6.

hence, the common ratio is 6.

Answer from: Quest

Here, feel free to ask if there are any questions

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problems 11 through 17
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